1 digit numbers : 0-9, => '5' appears only once = 1
2 digit numbers : 10-99, => Numbers with '5' at ten's place = 10 (50-59)
Numbers with '5' at unit's place = 9 (15,25,....,95) - '1' which is in 55' = 8
=> Total number of times digit '5' appears in the number from 1 to 100 = $$1+10+8=19$$
=> Ans - (C)
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